Calculate the molarity of an hcl solution if a 10.00 ml sample requires 25.24 ml of a 1.600 m naoh solution to be neutralized.

10.00 mL of 4.038 M HCl required 25.24 mL of 1.600 M NaOH to be neutralized.

Let's consider the following neutralization equation.

NaOH + HCl ⇒ NaCl + H₂O

A 10.00 mL (Va) sample of HCl requires 25.24 mL (Vb) of a 1.600 M (Cb) NaOH to be neutralized. Since the molar ratio of NaOH to HCl is 1:1, we can calculate the concentration of the acid (Ca) using the following expression.

10.00 mL of 4.038 M HCl required 25.24 mL of 1.600 M NaOH to be neutralized.

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The molarity of the HCl solution should be 4.04 M

Explanation:

Step 1: Data given

volume of HCl solution = 10.00 mL = 0.01 L

volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L

Step 2: The balanced equation

HCl + NaOH → NaCL + H2O

Step 3: Calculate molarity of HCl

n1*C1*V1 = n2*C2*V2

Since the mole ratio for HCl and NaOH is 1:1 we can just write:

C1*V1 =C2*V2

⇒ with C1 : the molarity of HCl = TO BE DETERMINED

⇒ with V1 = the volume og HCl = 10 mL = 0.01 L

⇒ with C2 = The molarity of NaOH = 1.6 M

⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L

C1 * 0.01 = 1.6 * 0.02524

C1 = (1.6*0.02524)/0.01

C1 = 4.04M

The molarity of the HCl solution should be 4.04 M

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